\documentclass[a4paper]{article}
\usepackage[margin=1in]{geometry}
\usepackage{ctex}
\usepackage{tikz}
\usepackage{color}
\usepackage{amssymb}
\usepackage{amsmath}
\usepackage{xltxtra}
\usepackage{mflogo,texnames}
\usepackage{graphicx}
\usepackage{mathpazo}
\usepackage{pgfplots}
\pgfplotsset{compat=newest}% use newest version

\newcommand*{\A}{2}
\newcommand*{\num}{2.2}

 % X 的参数方程
\pgfmathdeclarefunction{SolutionX}{1}{%
    \pgfmathparse{\A*(cos(deg(\t)))^3}%
}

 % Y 的参数方程
\pgfmathdeclarefunction{SolutionY}{1}{%
    \pgfmathparse{\A*(sin(deg(\t)))^3}%
}

 % define elegant style
\tikzset{elegant/.style={smooth, red, thick, samples=1001}}
\usetikzlibrary{arrows.meta}

\title{\heiti\zihao{2} 习题8.2}
\author{中书君}
\date{\songti 2021年1月13日}

\begin{document}
\maketitle
\section{求由曲线 $y=x$ 和 $y=x^{2}$ 所围成的平面图形绕 $x$ 轴旋转一周而成的旋转体体积.}
\textbf{解}\quad
\begin{tikzpicture}
    % draw the axis
   \draw[->] (-3,0) -- (3,0) node[below] {$x$};
   \draw[->] (0,-3) -- (0,3) node[above] {$f(x)$};
    % draw the function (piecewise)
   \draw[thin,domain=-1:3,samples=1000] plot(\x,\x);
   \draw[thin,domain=-1:3,samples=1000] plot(\x,\x^2);
\end{tikzpicture}
$$
V = \pi \int_0^{1}(x^2 - x^4) dx= \frac{2 \pi}{15}
$$

\section{求由 $y=x^{3}, y=8$ 及 $y$ 轴所围成的曲边梯形绕 $y$ 轴旋转一周而成的立体的体和.}
\textbf{解}\quad
\begin{tikzpicture}
    % draw the axis
   \draw[->] (-3,0) -- (3,0) node[below] {$x$};
   \draw[->] (0,-3) -- (0,3) node[above] {$f(x)$};
    % draw the function (piecewise)
   \draw[thin,domain=-1:1,samples=1000] plot(\x,1);
   \draw[thin,domain=-1:1,samples=1000] plot(\x,\x^3);
\end{tikzpicture}
$$
V=\pi \int_{0}^{8} y^{\frac{2}{3}} \mathrm{~d} y=\frac{160 \pi}{3}
$$


\section{求由旋轮线 $x=a(t-\sin t), y=a(1-\cos t), 0 \leqslant t \leqslant 2 \pi, a>0$ 与 $x$ 轴所围图形绕 $x$轴旋转所得旋转体的体积.}
\textbf{解}\quad
$$
V=\pi \int_{0}^{2 \pi} y^{2} d x=\pi \int_{0}^{2 \pi} a^{2}(1-\cos t)^{2} a(1-\cos t) d t=a^{3} \pi \int_{0}^{2 \pi}(1-\cos t)^{3} d t=5 \pi
$$


\section{求由星形线 $x=a \cos ^{3} t, y=a \sin ^{3} t$ 所围平面图形绕 $x$ 轴旋转所得立体的体积.}
\textbf{解}\quad
\begin{tikzpicture}
    \begin{axis}[axis lines=middle,
                 xmin=-\num, xmax = \num,
                 ymin=-\num, ymax = \num,
                 ylabel=$f(x)$,
                 xlabel=$x$]
        \addplot[elegant,variable=\t, domain=-2*pi:0]
            ({SolutionX(\t)},{SolutionY(\t)});
    \end{axis}
\end{tikzpicture}
$$
V=\pi \int_{-a}^{a} y^{2} \mathrm{~d} x=3 \pi a^{3} \int_{0}^{\pi} \sin ^{7} t \cos ^{2} t \mathrm{~d} t=6 \pi a^{3} \int_{0}^{\pi}\left(\sin ^{7} t-\sin ^{9} t\right) \mathrm{d} t=\frac{32 \pi a^{2}}{105}
$$


\section{求由 $(x-2)^{2}+y^{2}=1$ 所围平面图形绕 $y$ 轴旋转所得旋转体的体积.}
\textbf{解}\quad
此为一个以$(2,0)$为圆心，以$1$为半径的实心圆，可将其视为左半圆和右半圆。由于要求绕$y$轴转动，从而需要将$x$用$y$表示。
$x=2+\sqrt{1-y^2}$或$x=2-\sqrt{1-y^2}$。从而有：
$$
\begin{aligned}
V&=\pi \int_{-1}^{1}\left[\left(2+\sqrt{1-y^{2}}\right)^{2}-\left(2-\sqrt{1-y^{2}}\right)^{2}\right] \mathrm{d} y\\&
=8 \pi \int_{-1}^{1} \sqrt{1-y^{2}} \mathrm{~d} y=8 \pi \int_{\pi}^{2 \pi}-\sin ^{2} t \mathrm{~d} t=4 \pi^{2}
\end{aligned}
$$


\section{求平面图形 $y=\sin x(0 \leqslant x \leqslant \pi)$ 绕 $y$ 轴旋转所得旋转体的体积.}
\textbf{解}\quad
可将其分为$[0,\frac{\pi}{2}],[\frac{\pi}{2},\pi]$两部分。

当$0<x<\frac{\pi}{2}$时：
$$
V_{1}=\pi \int_{0}^{1}(\arcsin y)^{2} \mathrm{~d} y=\pi \int_{0}^{\frac{\pi}{2}} u^{2} \cos u \mathrm{~d} u=\frac{3 \pi^{2}}{4}-2 \pi, 
$$

当$\frac{\pi}{2}<x<\pi$时：
$$
V_{2}=\pi \int_{0}^{1}\left(\arccos y+\frac{\pi}{2}\right)^{2} \mathrm{~d} y=\pi \int_{\frac{\pi}{2}}^{0}\left(-t^{2}+\pi t\right) \sin t \mathrm{~d} t=2 \pi^{2}-2 \pi
$$

从而有:
$$
V_{2}-V_{1}=2\pi^{2}-\frac{\pi^3}{4}
$$


\section{过坐标原点 (0,0) 作曲线 $y=\ln x$ 的切线，该切线与曲线 $y=\ln x$ 以及 $x$ 轴所围成的 平面图形记为 $D$,计算 $D$ 绕 $x$ 轴旋转一周所得旋转体的体积.}
\textbf{解}\quad
切线为$y=\frac{x}{e}$，切点为$(e,1)$。从而有：
$$
S=\pi \int_{0}^{e} \frac{x^{2}}{e^{2}}dx-\pi \int_{1}^{e} \ln ^{2} xdx=\frac{e\pi}{3}-(e-2)\pi=2\pi-\frac{2e\pi}{3}
$$


\section{求心形线 $r=a(1-\cos \theta)(a>0)$ 所围区域绕极轴旋转一周所得立体的体积.}
\textbf{解}\quad
\begin{tikzpicture}[samples=200]
    \draw[-Stealth](-0.5,0)--(2.5,0)node[below]{$x$};
    \draw[-Stealth](0,-1.5)--(0,1.5)node[left]{$y$};
    \draw[domain=-pi:pi]plot({1.6*cos(\x/2 r)*cos(\x r)},{1.6*cos(\x/2 r)*sin(\x r)});
    \node at(-5pt,-5pt){$O$};\node[below]at(1.8,0){$2a$};
    \node at(1.5,-1.4){$r=a(1+\cos\theta)$};
\end{tikzpicture}
$$x=a(\cos\theta - \cos^2 \theta),y=a(\sin \theta - \sin\theta \cos\theta)$$


\section{计算椭球体 $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}+\frac{z^{2}}{c^{2}} \leqslant 1$ 的体积 $($ 利用已知截面面积的立体体积的求法).}
\textbf{解}\quad
椭圆面积公式：$S=\pi ab$
$$
V=\int_{-a}^{a} \pi \sqrt{1-\frac{x^{2}}{a^{2}}} b c \mathrm{~d} x=\frac{4}{3} \pi a b c
$$


\section{求下列平面曲线绕指定轴旋转一周所得旋转曲面的面积:}
\subsection{$y=\sin x, 0 \leqslant x \leqslant \pi,$ 绕 $x$ 轴}
\textbf{解}\quad
$$
S=2 \pi \int_{0}^{\pi} \sqrt{\cos ^{2} x+1} \sin x \mathrm{~d} x=2 \pi \int_{-1}^{1} \sqrt{t^{2}+1} d t=2 \pi \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec ^{3} u d u\\=2\pi\left(\left.\frac{\sec ^{2} u \sin u}{2}\right|_{-\frac{\pi}{4}} ^{\frac{\pi}{4}}+\frac{1}{2} \int_{-\frac{\pi}{4}}^{\frac{\pi}{4}} \sec u d u\right)=2\sqrt{2}\pi+2\pi\ln(\sqrt{2}+1)
$$



\subsection{$y=\sqrt{3-x^{2}},-1 \leqslant x \leqslant 1,$ 绕 $x$ 轴}
\textbf{解}\quad
$$
S=2 \pi \int_{-1}^{1} \sqrt{3-x^{2}} \cdot \sqrt{1+\frac{x^{2}}{3-x^{2}}} \mathrm{~d} x=2 \pi \int_{-1}^{1} \sqrt{3} \mathrm{~d} x=4 \sqrt{3} \pi
$$



\subsection{$\frac{x^{2}}{4}+y^{2}=1,$ 绕 $y$ 轴}
\textbf{解}\quad
令$x = 2\cos \theta,y = \sin \theta $

$$
S=2 \pi \int_{-\frac{\pi}{2}}^{\frac{\pi}{2}} 2 \cos \theta \sqrt{1+3 \sin ^{2} \theta} d \theta=4 \pi\left(\frac{\ln \left(\frac{2+\sqrt{3}}{2-\sqrt{3}}\right)}{2 \sqrt{3}}+2\right)
$$



\subsection{$x^{2}=4 y, 0 \leqslant x \leqslant 2,$ 绕 $y$ 轴}
\textbf{解}\quad
$$
S=2 \pi \int_{0}^{1} 2 \sqrt{y} \cdot \sqrt{1+\frac{1}{y}} d y=4 \pi \int_{1}^{2} \sqrt{y+1} d y+1=\frac{(16 \sqrt{2}-8) \pi}{3}
$$




\section{已知曲线 $L: y=\sqrt{x-1},$ 过坐标原点 (0,0) 作曲线 $L$ 的切线,求由该切线、曲线 $L$ 以 及 $x$ 轴所围成的平面图形绕 $x$ 轴旋转一周所得旋转曲面的面积.}
\textbf{解}\quad
切线：$y=\frac{x}{2}$
$$
\begin{aligned}
&S_{1}=2 \pi \int_{0}^{2} \frac{x}{2} \cdot \sqrt{1+\frac{1}{4}} \mathrm{~d} x=\sqrt{5} \pi\\&
S_{2}=2 \pi \int_{1}^{2} \sqrt{x-1} \sqrt{1+\frac{1}{4(x-1)}} \mathrm{~d} x=\frac{5 \sqrt{5}-1}{6} \pi\\&
S=S_{1}+S_{2}=\frac{11 \sqrt{5}-1}{6} \pi
\end{aligned}
$$


\section{求旋轮线 (摆线)的一拱 $\left\{\begin{array}{l}x=a(t-\sin t), \\ y=a(1-\cos t),\end{array} 0 \leqslant t \leqslant 2 \pi\right.$ 绕 $x$ 轴旋转一周所得旋转曲面的面积.}
\textbf{解}\quad
$$
\begin{aligned}
S&=2 \pi \int_{0}^{2 \pi} y(t) \sqrt{x^{\prime2}(t)+y^{\prime2}(t)} d t= 2 \pi \int_{0}^{2 \pi} a(1-\cos t) \sqrt{a^{2}(1-\cos t)^{2}+a^{2} \sin ^{2} t} d t
\\&=2 \pi \int_{0}^{2 \pi} \sqrt{2} a^{2}(1-\cos t)^{\frac{3}{2}} \mathrm{~d} t
=\frac{64 \pi a^{2}}{3}
\end{aligned}
$$


\section{求下列曲线绕极轴旋转所得旋转曲面的面积:}
\subsection{心形线 $r=a(1-\cos \theta),(a>0) $}
\textbf{解}\quad
$$
\begin{aligned}
 \mathrm{S} &=2 \pi \int_{0}^{\pi} r \sin \theta \sqrt{r^{2}+r^{\prime 2}} \mathrm{~d} \theta=4 a^{2} \pi \int_{0}^{\pi} \sin \theta(1-\cos \theta) \sin \frac{\theta}{2} \mathrm{~d} \theta \\
&=16 a^{2} \pi \int_{0}^{\pi} \sin ^{4} \frac{\theta}{2} \cos \frac{\theta}{2} \mathrm{~d} \theta=\frac{32 \pi a^{2}}{5}
\end{aligned}
$$

\subsection{双纽线 $r^{2}=a^{2} \cos 2 \theta,(a>0)$}
\textbf{解}\quad
$$
\mathrm{S}=4 \pi \int_{0}^{\frac{\pi}{4}} r \sin \theta \sqrt{r^{2}+r^{\prime 2}} \mathrm{~d} \theta=4 \pi a^{2} \int_{0}^{\frac{\pi}{4}} \sin \theta \mathrm{d} \theta=(4-2 \sqrt{2}) \pi a^{2}
$$






\end{document}